Calculus 3 First Half Flashcards
What does eliminating the parameter do?
Turns a parametric equation that has x = ƒ(t) and y = ƒ(t) into just a relationship between x
and y
Eliminate the Parameter
x = 2t-4 , y = 4t^2
y = (x+4)^2
sin(t)^2 + cos(t)^2 = ?
1
Eliminate the Parameter and then find the x and y values within the parameter.
(hint: uses circle equation form)
x = 3sin(t) , y = 3cos(t), 0 ≤ t ≤ π
(x/3)^2 + (y/3)^2 = 1
-> x^2 + y^2 = 9
t = 0, π/2 , π , 3π/2 , 2π
x = 0, 3, 0, -3, 0
y = 3, 0, -3, 0, 3
Derivative of Parametric Function
Dy/Dx = (d/dy) / (d/dx)
What does the derivative of a parametric give you?
It will tell you the slope of the tangent line.
The slope of a tangent line
y-y0 = m(x-x0)
When is there a verticle tangent in a parametric?
where dx/dt = 0
or where dy/dt ≠ 0
Where are the critical points of a parametric equation?
Where dy/dx = 0
What are the only variables that should be in your dy/dx once you have eliminated the parameters?
t
How do you find the X and Y values of the critical points in a parametric equation?
After setting your dy/dx equal to zero and finding you ‘t’ value, plug that t value back into your x(t) and y(t) equations.
What tells you the concavity of a curve?
The second derivative. When 2nd derivative > 0 , then it is concave up. When < 0 , it is concave down
How to find the area underneath the parametric curve?
Integral from a to b (y(t) x’(t)) dt
Cartesian Coordinates
(x,y)
Polar Coordinates
(r,ø)
